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Author Topic: Rolling Resistance  (Read 24127 times)
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thecarfarmer
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« Reply #30 on: August 18, 2012, 11:40:00 AM »

The formula I have is N = W x V x T / 270 where N is the power required, W is the weight in kilograms, V is the speed in kilometers per hour, and T is the rolling resistance of the tire.
 (snip)

Units for "N"?  Newtons or KWH, or something?

Thanks!

-Bill
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Jack Gifford
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« Reply #31 on: August 18, 2012, 11:40:08 PM »

You'll need Ratliff to answer about the formula he posted. To utillize his formula, you'll also need (besides units for N) units for T.

The units for N must represent power; Newton is the MKS unit of force, not power, and KWH is a unit of energy, not power.
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PatMc
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« Reply #32 on: August 19, 2012, 01:40:58 PM »

One thing I've never got a solid answer out of, is if LSR tires slip.  Slip is not spinning.  It's the stretching of the rubber.  With no loss of traction, a tire can move the vehicle less than it's circumference when torque is applied.   This is normal on non-LSR tires.

Think of a centipede crawling.  Tires do the same thing when torque is applied.
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hotrod
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« Reply #33 on: August 19, 2012, 03:11:12 PM »

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The units for N must represent power; Newton is the MKS unit of force, not power, and KWH is a unit of energy, not power.

No rolling resistance is a force, to compute the power needed to over come it you need to know how fast the car is going.
N = newtons the force resisting motion.

There are several other rolling resistance formulas out there in various engineering texts and some books on aerodynamics, mention it as well.
The best way to determine it is to use low speed coast down information where the dominant form of drag is rolling resistance rather than aerodynamic drag.

Larry
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Jack Gifford
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« Reply #34 on: August 20, 2012, 12:27:56 AM »

Talk about muddying the water...
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Harold Bettes
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« Reply #35 on: August 20, 2012, 03:48:21 PM »

A Newton is equal to .2248089lbs force. Or, 1lbf = 4.448222 Newtons. rolleyes

And of course, one needs to consider a sign change for resistance in calculations. tongue

Once you get started in force calculations for a salt rider, lots of things are happening (in motion) at the same time. grin

It is however much more simple than attempting to keep up with the National Debt Clock. shocked

Regards to All,
HB2 smiley
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wobblywalrus
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« Reply #36 on: August 20, 2012, 07:43:56 PM »

The best we can do with this complicated situation is to collect data from our runs and to develop slip coefficients for the combination of driver and vehicle.  There are so many things happening and one of the most significant is the driver/rider.  Some people know how to go fast by minimizing the application of excess power and the resultant slip.  Others do the opposite.
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Old Scrambler
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« Reply #37 on: August 24, 2012, 08:03:39 PM »

About 2 years ago I started making plans to build a bike for the salt-flats. I knew it would be a Triumph Cub with about 20 hp and rather light, yet it had to carry my 200 lbs. of body weight.  I did not consider the amount of energy or power I would need to overcome rolling resistence. Rather, I thought about selecting the lightest, roundest, narrowest, firmest, and most reliable tires and wheels available within my almost nothing budget. From a practical point of view, I know that once the wheel is put into motion, a larger diameter is easier to push or pull with my own strength, such as a wheelbarrow or wagon. Practical experience seemes to work. I can almost effortlessly push my bike across the salt with one hand and when I stop I actually have to reverse my effort on the bike if I want it to stop with me. That's my formula......Oh, BTW.....how many minutes does your wheel keep spinning when suspended after you give it a good start ?
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